Answer: the pH of the solution is 4.52
Explanation:
Consider the weak acid as Ha, it is dissociated as expressed below
HA H⁺ + A⁻
the Henderson -Haselbach equation can be expressed as;
pH = pKa + log( [A⁻] / [HA])
the weak acid is dissociated into H⁺ and A⁻ ions in the solution.
now the conjugate base of the weak acid HA is
HA(aq) {weak acid} H⁺(aq) + A⁻(aq) {conjugate base}
so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;
pKₐ = -logKₐ
pKₐ = -log ( 7.4×10⁻⁵ )
pKₐ = 4.13
now thw pH is
pH = pKₐ + log( [A⁻] / [HA])
pH = 4.13 + log( [0.540] / [0.220])
pH = 4.13 + 0.3899
pH = 4.5199 = 4.52
Therefore the pH of the solution is 4.52
There are things called "Reactants" and "Products" All chemical equations look something like "A + B →C (+ D...)," in which each letter variable is an element or a molecule (a collection of atoms held together by chemical bonds). The arrow represents the reaction or change taking place. Some equations may have a double-headed arrow (↔), which indicates that the reaction can proceed either forward or backward. When a compound has been written out, you must identify the elements and know their chemical symbols. The first element written is “first name” of the compound. Use the periodic table to find the chemical symbol for the element. So here is an example: Dinitrogen hexafluoride. The first element is nitrogen and the chemical symbol for nitrogen is N. To know the numbers of atoms that are present for each element you can just look at the prefix from the element For example: Dinitrogen has a the prefix “di-“ which means 2; therefore, there are 2 atoms of nitrogen present.
Write dinitrogen as N2.
Now for the second element or "last name" of the compound whatever will follow the first element so like; Dinitrogen hexafluoride. The second element is fluorine. Simply replace the “ide” ending with the actual element name. The chemical symbol for fluorine is F.
But the more you practice with, the easier it will be to decipher chemical formulas in the future and learn the language of chemistry.
Sulfur dioxide: SO2
Carbon tetrabromide: CBr4
Diphosphorus pentoxide: P2O5 ← That is one of the examples I'll give you.
have a gooooood daaaaayy
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
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